Variable name has a variable at the end

Question

if (playerturn == 1) {
     currentpos1 = startpos1 + dicetotal;
     prevpos1 = currentpos1 - dicetotal;
 } else {
     currentpos2 = startpos2 + dicetotal;
     prevpos2 = currentpos2 - dicetotal;
 }

I know this is a slight duplicate of other questions, but I still didn't understand the answers. This is for a monopoly board game, which can have upto 6 players. I want to clean up the code, because the only thing that changes is the last part of the variable.

Currentpos(i), Prevpos(i), Startpos(i) are variable names. So how would I concatenate a variable into a variable name?

Quentin ok. Perhaps you could show me how I could use arrays or objects instead? — Elton Frederik, Dec 25 '13 at 10:36
And no my question is not a duplicate. They are trying to have variable variables. Whereas I am trying to make 6 different variables, but want to automate naming them by adding a variable at the end of their names — Elton Frederik, Dec 25 '13 at 10:40

nzn · Accepted Answer · 2013-12-25T11:10:28.573

2

You can use object instead of variables.

var players = {
      playername : { 
          currpos:'val',
          prevpos:'val',
          startpos:'val',
          ....
      },
      ....

  }

Adding new player

players['playername1'] = { 
          currpos:'val',
          prevpos:'val',
          startpos:'val',
          ....
      }

And usage is players['playername'].currpos so in yur case jus players['playername'+playerturn].currpos = players['playername'+playerturn].startpos + dicetotal;

or array

var players = [
    { 
        currpos:'val',
        prevpos:'val',
        startpos:'val',
        ....
    },
    { 
        currpos:'val',
        prevpos:'val',
        startpos:'val',
        ....
    }
];

Adding new player:

 players.push({ 
        currpos:'val',
        prevpos:'val',
        startpos:'val',
        ....
    });

And usage is players[0].currpos so in your case players[playerturn].currpos = players[playerturn].startpos + dicetotal

edited Dec 25 '13 at 11:10

answered Dec 25 '13 at 10:57

nzn

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Thanks nzn. I will play around with it and see if I can get it to work. – Elton Frederik Dec 25 '13 at 10:59
@EltonFrederik, even though the answer seems correct, you shouldn't mark it as a correct answer until you are absolutely sure of it. – anar khalilov Dec 25 '13 at 11:26
Yes Anar you are right, I got carried away. But anyway I can confirm now that I got it to work. So all is good. – Elton Frederik Dec 25 '13 at 14:45

Variable name has a variable at the end

1 Answers1